Given a non-empty, singly linked list with head node head, return a middle node of linked list.

If there are two middle nodes, return the second middle node.

Example 1:

Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3. (The judge’s serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.

Example 2:

Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.

方法1：链表长度

从头指针遍历链表得到链表长度，然后找到中间。此处对于偶数长度的链表返回后面一个结点对象，因而可以直接返回cur.next。如果要返回前面一个结点对象，则需要对奇偶情况分别输出。

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode middleNode(ListNode head) {
       ListNode cur = head;
        int count = 0;
        while (cur != null) {
            count++;
            cur = cur.next;

        }
        cur = head;
        for (int i = 1; i <= count / 2; i++) {
            cur = cur.next;
        }
        return cur.val;
    }
}

方法二：双指针

fast和slow快慢双指针，fast每次走两步，slow每次走一步，偶数长度的链表fast.next==null说明已经走到尾部，奇数长度的链表fast.next.next == null说明已经走到尾部。

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class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        if (fast.next == null) {
            return slow;
        } else return slow.next;
    }
}